Is a core for the generator of a Feller semi-group invariant under the resolvent? Let $\{T_t:t\geq 0\}$ be a Feller semi-group acting on $C_0(\mathbb{R})$ with generator $(A,\mathcal{D}_A)$. We know a subspace $D\subset \mathcal{D}_A$ is a core for $A$ if $(\lambda-A)D$ is dense in $C_0(\mathbb{R})$ for some $\lambda>0$. Another condition for a subspace $D\subset \mathcal{D}_A$ to be a core for $A$ is that it be dense in $C_0(\mathbb{R})$ and invariant under the semi-group, i.e. $T_t(D)\subset D$ for all $t\geq 0$. For $\alpha>0$, let $R_{\alpha}:C_0(\mathbb{R})\to \mathcal{D}_A$ be the corresponding resolvent operator, i.e. $R_{\alpha}=(\alpha-A)^{-1}=\int_0^{\infty}\exp(-\alpha t)T_t dt$. Notice that $\mathcal{D}_A$ is a core for $A$ and that $R_{\alpha}\mathcal{D}_A\subset \mathcal{D}_A$. My question is whether this invariance under the resolvent holds for every core or perhaps there is some easy to check condition that implies resolvent invariance in certain cases.
 A: In general, cores fail to be invariant under the resolvent (see the example below). However, invariance under the resolvent is a sufficient condition for a set to be a core. More precisely:

Let $(T_t)_{t \geq 0}$ be a Feller semigroup with generator $(A,\mathcal{D}_A)$ and resolvent $(R_{\alpha})_{\alpha>0}$. If $\mathcal{D} \subseteq C_0(\mathbb{R})$ is a dense subset such that $R_{\alpha}(\mathcal{D}) \subseteq \mathcal{D}$ for some $\alpha>0$, then $\mathcal{D}$ is a core for the generator $(A,\mathcal{D}_A)$.

For a proof see e.g. Lévy Matters III by Böttcher, Schilling & Wang, Lemma 1.34.
The following example shows that invariance under the resolvent is not necessary for being a core:
Example: Let $(N_t)_{t \geq 0}$ be a Poisson process with intensity $1$. The associated semigroup is given by
$$T_t f(x) := \mathbb{E}f(x+N_t), \qquad f \in C_0(\mathbb{R}), t \geq 0, x \in \mathbb{R},$$
and it is not difficult to see that $(T_t)_{t \geq 0}$ is a Feller semigroup acting on $C_0(\mathbb{R})$ with generator
$$Af(x) = f(x+1)-f(x), \qquad f \in \mathcal{D}_A := C_0(\mathbb{R}).$$
Clearly, the space of smooth functions with compact support $C_0^{\infty}(\mathbb{R})$ is a core for the generator $(A,C_0(\mathbb{R}))$. Now let $f \in C_0^{\infty}(\mathbb{R})$ be a non-negative function such that $f(0)>0$. Since $\mathbb{P}(N_t=k)>0$ for any $k \in \mathbb{N}$ and $t>0$ we have
$$(R_{\alpha} f)(-k) = \int_0^{\infty} e^{-\alpha t} \mathbb{E}f(-k+N_t) \, dt \geq f(0) \int_0^{\infty} e^{-\alpha t} \mathbb{P}(-k+N_t=0) \, dt >0$$
for any $\alpha>0$ and $k \in \mathbb{N}$. This shows that $R_{\alpha} f$ does not have compact support; in particular, $R_{\alpha} f \notin C_0^{\infty}(\mathbb{R})$. This proves that the core $\mathcal{D} := C_0^{\infty}(\mathbb{R})$ does not satisfy the relation $R_{\alpha} \mathcal{D} \subseteq \mathcal{D}$.
A: I think the following is a counter example.
Take the Feller semigroup $T=\{T_t\}_{t\ge 0}$ of a negative drift stopped at $0$,  defined as 
\begin{equation}
T_tf(x):=\left\{\begin{split}
f(x-t), & \quad &x>t,\\
f(0), &\quad  &x\le t,
\end{split}
\right.
\end{equation}
on the space $C_0([0,\infty))$ of real-valued continuous functions on $[0,\infty)$ vanishing at infinity equipped with the supremum norm. Then the generator is
$$
Af(x):=\left\{\begin{split}
-\partial_xf(x), & \quad &x>0,\\
0, &\quad  &x=0,
\end{split}
\right.
$$
with the domain 
$$
\mathcal D_A =\{f\in C^1((0,\infty)),\ \partial_xf(x)\in C_0((0,\infty)),\ \partial_xf(0+)=0\}
$$
Take $$
D:= \{f\in \mathcal D_A: \ \partial_{xxx}f(0+)=0, \text{ derivatives up to the third exist and are in } C_0([0,\infty))\}.
$$ Observe that the subspace $D$ is dense in $\mathcal D_A$ and invariant under the semigroup $T$, hence it is a core for $(A,\mathcal D_A)$.
Pick any $f\in D$ such that $\partial_{xx}f(0+)\neq 0$, fix $\alpha>0$ and compute $\partial_{xxx}R_\alpha f(0+)$. For $x>0$
\begin{align}
\partial_{xxx}R_\alpha f(x) &= \partial_{xxx}\left(\int_0^x e^{-\alpha t}f(x-t)dt+f(0+)\alpha^{-1}e^{-\alpha x}\right)\\
 &= \partial_{xx}\left(\int_0^x e^{-\alpha t}\partial_xf(x-t)dt\right)\\
 &= \partial_{x}\left(\int_0^x e^{-\alpha t}\partial_{xx}f(x-t)dt+ e^{-\alpha x}\partial_xf(0+)\right)\\
 &= \partial_{x}\left(\int_0^x e^{-\alpha t}\partial_{xx}f(x-t)dt\right)\\
 &=\int_0^x e^{-\alpha t}\partial_{xxx}f(x-t)dt+e^{-\alpha x}\partial_{xx}f(0+).
\end{align}
Then $\lim_{x\downarrow 0}\partial_{xxx}R_\alpha f(x) = \partial_{xx}f(0+)\neq 0$, which implies that $R_\alpha f \notin D$. $\quad\square$ 
Observe that the core
$$
\tilde D:= \{f\in \mathcal D_A: \ \partial_{xx}f(0+)=0, \text{ derivatives up to the second exist and are in } C_0([0,\infty))\}
$$
would be invariant under $R_\alpha$, which makes me suspect that the invariance of the core under the resolvent is generally hard to check at least on bounded domains.
